Given a horizontal mass-spring system on a frictionless surface, derive the equation of motion (position as a function of time) for the mass with the following initial conditions: $$x(0)=1 cm$$ and $$\dot{x}(0)=0 \frac{m}{s}$$. The spring has a constant of $$2 \frac{N}{m}$$ and mass has $$m=500 g$$.

Justin A.

Answer:

Newton tells us that $$F_{net}=ma=m\ddot{x}$$
The only force acting on the mass after release is the spring’s restoring force:
$$F_{net}=F_{rest}=-kx$$
(The restoring force is negative in order to always oppose the direction of motion of the mass [i.e. if the mass is pulled to the right, the force will apply to left. If the mass is pushed into the left, the force will apply to the right].)
So:
$$-kx=m\ddot{x}$$
$$\ddot{x}+\frac{k}{m} x=0$$
This is a second order, linear, ordinary differential equation with constant coefficients. The characteristic polynomial is:
$$r^2+\frac{k}{m}=0→r^2=-\frac{k}{m}→r=±\sqrt{\frac{k}{m}} i$$
The solution has the form:
$$x(t)=c_1 cos⁡[\sqrt{\frac{k}{m}}t]+c_2 sin⁡[\sqrt{\frac{k}{m}}t]$$
Now apply the initial conditions to find the constants:
$$x(0)=0.01 m= c_1+0→c_1=0.01$$
So, we now have:
$$x(t)=0.01 cos⁡[\sqrt{\frac{k}{m}}t]+c_2 sin⁡[\sqrt{\frac{k}{m}}t]$$
Now, the second initial condition requires the derivative:
$$\dot{x}(t)=-0.01 \sqrt{\frac{k}{m}} sin⁡[\sqrt{\frac{k}{m}}t]+c_2 cos⁡[\sqrt{\frac{k}{m}}t]$$
$$\dot{x}(0)=0=0+c_2→c_2=0$$
This gives:
$$x(t)=0.01 cos⁡[\sqrt{\frac{k}{m}}t]$$
Plugging in the constants yields a final position as a function of time given the initial conditions of:
$$x(t)=0.01 cos⁡[\sqrt{\frac{2}{0.5}}t]=0.01 cos⁡[\sqrt{4}t]=0.01 cos[⁡2t]$$
$$x(t)=0.01 cos⁡ (2t)$$

Physics (Electricity and Magnetism)

TutorMe

Question:

a) Derive the equation for the magnitude of a magnetic field as a function of distance from a straight, infinitely long, constant current carrying wire. b) Using the equation from part a, determine the strength of the induced magnetic field $$2 cm$$ away from a wire carrying $$10 A$$ of constant current.

Justin A.

Answer:

a) Ampere’s Law:
$$ \oint{\vec{B}\cdot d\vec{s}} = \mu_0 I +\frac{1}{c^2} \frac{\partial}{\partial t} \int{\vec{E}\cdot d\vec{A}} $$
With the current not varying with time, the second term is zero due to the derivative:
$$ \oint{\vec{B}\cdot d\vec{s}} = \mu_0 I $$
Taking a circle about the wire at an arbitrary radial distance,$$ r$$, the length of the circle is just the circumference, as $$2\pi r$$. Additionally, by use of the right-hand rule, the B-field lines will always lie parallel to the taken circle at each point. The integral on the left-hand side becomes when we go to magnitude:
$$||{\vec{B}}|| \cdot 2\pi || \vec{r}|| cos (0)=\mu_0 I $$
$$ B(2\pi r) = \mu_0 I $$
$$ B=\frac{μ_0 I}{2\pi r} $$
b) Using the above equation and plugging in given values:
$$ B=\frac{(4\pi \times 10^{-7} N/A^2 )(10 A)}{(2π(0.02 m))}≈0.0001 N/(A\cdot m)≈0.1 mT$$

Physics

TutorMe

Question:

Two masses are in constant contact. The masses are $$ m_1=5 kg $$ and $$ m_2=2 kg $$. The larger mass is on the left and a force of $$ 20 N $$ is continuously applied to the left of the larger mass. a) What is the acceleration of the system and b) what is the force on mass one from mass two ( $$ F_{21} $$ )?

Justin A.

Answer:

a) System acceleration:
$$F_{net}=m_{sys} a_{sys}$$
$$20 N=(7 kg) a_{sys}$$
$$a_{sys}=\frac{20 N}{7 kg}=\frac{20}{7} \frac{kg\cdot m}{kg\cdot s^2}≈2.857 \frac{m}{s^2}→a_{sys}≈2.9 \frac{m}{s^2}$$
b) Interface force:
$$F_{12}=m_2 a_{sys}$$
$$F_{12}≈(2 kg)(2.9 \frac{m}{s^2} )≈5.8 N$$
Via Newton’s 3rd law – The force on mass two from mass one is equal and opposite to the force on mass one from mass two.
$$F_{12}=-F_{21}$$
$$F_{21}≈-5.8 N $$ or $$ F_{12}≈5.8 N$$
The force pair at the interface cancels each other out. Physically, this means the two masses stay touching as the entire system accelerates (there is no motion relative to one another, the sum of the forces at the interface is zero.)

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